Even, Odd & Skip-Counting — Free Math Kangaroo Lesson (Grades 1–4)

About this topic

Numbers love to make groups. Some make tidy pairs. Some always keep one little dot alone. Some hop in steps you can see coming a mile away.

In this lesson you will pair up dots, hop along a number line, color a hundred square, and share cookies. You do not need any big words. Look at the pictures and count along. The numbers will start to show you their secrets.

CHAPTER 1

Even and Odd

THEORY

Two friends always hold hands. Line up your dots that way — two by two, like socks.

Try it with a few numbers and watch what happens.

Did you feel the pattern? 2, 4, 6 pair up perfectly. 3 and 5 always leave one orange dot with no buddy.

Numbers that pair up with nobody left out are even. Numbers with one lonely dot are odd.

Now here is the lovely part. You do not have to draw dots every time. Just look at the last digit:

Even numbers end in 0, 2, 4, 6, or 8. Odd numbers end in 1, 3, 5, 7, or 9. The big number can be huge — you still only peek at the last digit.

🎯 Try it

You have 8 dots. You put them in pairs, two by two. How many pairs do you get?

Here is how: 2 and 2 and 2 and 2 makes 4 pairs. No dot is left alone, so 8 is even.

THE TRICK

THE MOVE: Do not count the whole number. Peek at the last digit. Ends in 0, 2, 4, 6, or 8? Even. Ends in 1, 3, 5, 7, or 9? Odd.

WATCH OUT

Zero is even! 0 dots make zero pairs, and there is nobody left alone. So a number ending in 0 is always even, never odd.

WORKED EXAMPLE

PROBLEM · 2013 #11

Daniel had 36 sweets. He shared them equally between his siblings. How many siblings can he definitely not have?

A) 2 B) 3 C) 4 D) 5 E) 6

Daniel has 36 sweets to share with his brothers and sisters, the same number to each, none left over. We want the number of siblings he can NOT have.

So we ask one little question over and over: can 36 split into equal piles with nobody left out?

2 piles: 18 and 18. Yes. 3 piles: 12 each. Yes. 4 piles: 9 each. Yes. 6 piles: 6 each. Yes.

Now 5 piles: 5 and 5 and 5... gets you to 35, with 1 sweet stuck alone. It does not share evenly. So he cannot have 5 siblings.

I keep asking the same tiny question: does it split evenly, with nobody left out? Four of the choices work. Five is the one that leaves a sweet alone — that is the answer.

Answer: D — 5

RULE OF THUMB

Even = pairs up, none alone. Odd = one left alone. Check the last digit and you are done.

MORE LIKE THIS

2020 · #16 Grandma has just baked 23 cupcakes and wants to give the same number of them to each of her six grandchildren, eating what is left over....

Grandma has just baked 23 cupcakes and wants to give the same number of them to each of her six grandchildren, eating what is left over. At least how many cupcakes will she have left to eat?

Show answer

Answer: E — 5

Show hints

Hint 1 of 2

Give each of the 6 children the same number of cupcakes, as many as you can.

Still stuck? Show hint 2 →

Hint 2 of 2

Whatever does not fit into an equal share is what Grandma eats.

Show solution

Approach: share equally, then see what is left over

Give each child 3 cupcakes: that uses 6 × 3 = 18 cupcakes.
Giving each a 4th would need 24, but there are only 23, so 3 each is the most that shares evenly.
That leaves 23 − 18 = 5 cupcakes for Grandma to eat.

2013 · #5 Five children are talking about the number 325. Andreas: “It is a three-digit number.” Boris: “All the digits are different.” Sara: “The...

Five children are talking about the number 325. Andreas: “It is a three-digit number.” Boris: “All the digits are different.” Sara: “The digit sum is 10.” Gerda: “The units digit is 5.” Daniela: “All the digits are odd.” Who has made a mistake?

Show answer

Answer: E — Daniela

Show hints

Hint 1 of 2

Check each child's statement against the actual number 325.

Still stuck? Show hint 2 →

Hint 2 of 2

One statement is simply false — look at whether every digit is really odd.

Show solution

Approach: test each claim about 325

325 is a three-digit number (Andreas correct), its digits 3, 2, 5 are all different (Boris correct).
3+2+5 = 10 (Sara correct) and the units digit is 5 (Gerda correct).
But 2 is even, so 'all the digits are odd' is wrong — that is Daniela.

CHAPTER 2

Skip-Counting and Multiples

THEORY

You already skip-count without knowing it. Count your fingers in twos: 2, 4, 6, 8, 10. That is skip-counting!

Skip-counting means you hop over numbers instead of stepping on every one. Watch the hops by 2 on a number line:

The numbers you land on — 2, 4, 6, 8, 10 — are the multiples of 2. A multiple is just a spot you land on when you skip-count.

Now color a hundred square. Paint every number you land on hopping by 5, and a pattern jumps out:

Every multiple of 5 ends in 5 or 0. Every multiple of 10 ends in 0. You can spot them from across the room, without counting.

🎯 Try it

Skip-count by 5: 5, 10, 15, 20, ? — what comes next?

Here is how: Each hop adds 5. After 20 comes 25. And look — it ends in 5, just like the pattern says.

THE TRICK

THE MOVE: To check a multiple of 5, look at the last digit. Ends in 5 or 0? It is a multiple of 5. Ends in 0? It is a multiple of 10 too.

WATCH OUT

30 ends in 0, so it is a multiple of 5 AND a multiple of 10. A number ending in 0 belongs to both clubs at once.

WORKED EXAMPLE

PROBLEM · 2017 #11

Balloons are sold in packages of 5, 10 or 25 pieces each. Marius buys exactly 70 balloons. What is the minimum number of packages he has to buy?

A) 3 B) 4 C) 5 D) 6 E) 7

Balloons come in packs of 5, 10, or 25. Marius needs exactly 70, with as few packs as he can.

Fewer packs means grab the biggest packs first. Take two packs of 25: that is 25 and 25, which is 50 balloons.

He still needs 70 − 50 = 20 more. The fastest 20 is two packs of 10.

Count the packs: 2 big ones and 2 tens = 4 packs.

Biggest packs first means the fewest packs. I keep grabbing the largest size that still fits, and skip-count up to 70.

Answer: B — 4

RULE OF THUMB

Multiples of 5 end in 5 or 0. Multiples of 10 end in 0. Multiples of 2 end in 0, 2, 4, 6, or 8.

MORE LIKE THIS

2013 · #14 In a shop you can buy oranges in bags of 4 or bags of 10. Pedro wants to buy exactly 48 oranges. What is the smallest number of bags he must buy?

In a shop you can buy oranges in bags of 4 or bags of 10. Pedro wants to buy exactly 48 oranges. What is the smallest number of bags he must buy?

Show answer

Answer: C — 6

Show hints

Hint 1 of 3

Fewer bags means use the big bags of 10 as much as you can.

Still stuck? Show hint 2 →

Hint 2 of 3

Try four bags of 10, which is 40 oranges, and see how many oranges are still needed.

Still stuck? Show hint 3 →

Hint 3 of 3

Fill the leftover with bags of 4, then count how many bags you used altogether.

Show solution

Approach: use the largest bags first

Four bags of 10 give 40, leaving 8, which is two bags of 4.
That is 4 + 2 = 6 bags, and no smaller number of bags reaches exactly 48.
So he needs 6 bags.

2011 · #9 A chicken farmer packs eggs in boxes of 6 and boxes of 12. What is the smallest number of boxes he needs to pack 66 eggs?

A chicken farmer packs eggs in boxes of 6 and boxes of 12. What is the smallest number of boxes he needs to pack 66 eggs?

Show answer

Answer: B — 6

Show hints

Hint 1 of 2

Use as many big boxes (of 12) as you can first.

Still stuck? Show hint 2 →

Hint 2 of 2

After the big boxes, fill the rest with boxes of 6.

Show solution

Approach: greedy with the larger box

66 = 12 × 5 + 6, so five boxes of 12 hold 60 eggs and one box of 6 holds the last 6.
That is 5 + 1 = 6 boxes, and no smaller number works, answer B.

★ MINI-QUIZ

Quick Check: Pairs and Hops

Two warm-ups about pairs and skip-counting. Take your time and draw if it helps.

2025 · #1 Pablo has six balloons. He gives away two of his balloons. How many balloons does Pablo have now?

Pablo has six balloons. He gives away two of his balloons. How many balloons does Pablo have now?

Show answer

Answer: C — 4

Show hints

Hint 1 of 2

You start with six and lose two of them.

Still stuck? Show hint 2 →

Hint 2 of 2

"Gives away" means take away — this is just subtraction.

Show solution

Approach: subtract what is given away

Pablo begins with 6 balloons.
He gives away 2, so take 2 away from 6.
6 − 2 = 4, so Pablo has 4 balloons.

2012 · #5 13 children play hide and seek. One of them is the seeker. After a little while 9 children have been found. How many are still hiding?

13 children play hide and seek. One of them is the seeker. After a little while 9 children have been found. How many are still hiding?

Show answer

Answer: A — 3

Show hints

Hint 1 of 2

One of the 13 is the seeker, so how many are actually hiding?

Still stuck? Show hint 2 →

Hint 2 of 2

Subtract the children already found from the number hiding.

Show solution

Approach: subtract in two steps

One child seeks, so 13 - 1 = 12 children are hiding.
Of those, 9 have been found, leaving 12 - 9 = 3.
So 3 children are still hiding.

CHAPTER 3

Sharing Equally and Leftovers

THEORY

It is snack time. You have cookies and some friends. The fair thing is: everyone gets the same. But sometimes the cookies do not split just right, and a few are leftover.

Share 7 cookies between 2 plates and see:

Each plate gets 3. One cookie is leftover. You cannot snap the last cookie into 2 whole cookies, so it just waits.

If there is nothing left over, the cookies share evenly. 6 cookies on 2 plates is 3 and 3, with none left.

🎯 Try it

You share 10 stickers between 3 friends, same amount each. How many stickers are leftover?

Here is how: Give each friend 3 stickers. That uses 9. You started with 10, so 1 is leftover.

THE TRICK

THE MOVE: Give out full rounds, the same to each friend, until you cannot give a full round anymore. Whatever is left over could not make one more round.

WATCH OUT

The leftover is always smaller than the number of friends. With 3 friends you can never have 3 or more leftover — you would just hand one more to everybody!

WORKED EXAMPLE

PROBLEM · 2020 #16

Grandma has just baked 23 cupcakes and wants to give the same number of them to each of her six grandchildren, eating what is left over. At least how many cupcakes will she have left to eat?

A) 1 B) 2 C) 3 D) 4 E) 5

Grandma has 23 cupcakes for 6 grandchildren, the same number each. She eats whatever is left over.

Skip-count by 6 to see how many full rounds fit: 6, 12, 18. The next one, 24, is too big — she only has 23.

So each child gets 3 cupcakes. That uses 6 × 3 = 18 cupcakes.

Left over: 23 − 18 = 5 cupcakes for Grandma.

I fill up rounds of 6 until I cannot do one more. Whatever is left is too small to share — that is what Grandma gets to eat.

Answer: E — 5

RULE OF THUMB

Shares evenly = nothing left over. Otherwise, the bit that does not fit one more round is the leftover.

MORE LIKE THIS

2017 · #13 13 children registered for a competition. Then another 19 joined. Six equally big teams are needed for the competition. How many more...

13 children registered for a competition. Then another 19 joined. Six equally big teams are needed for the competition. How many more children are needed, so that six equally big teams can be formed?

Show answer

Answer: D — 4

Show hints

Hint 1 of 2

First add the two groups of children together.

Still stuck? Show hint 2 →

Hint 2 of 2

Six equal teams means the total must split into 6 equal piles, so count up by sixes past your total.

Show solution

Approach: count up by sixes to the first number past the total

Altogether there are 13 + 19 = 32 children.
Six equal teams need a total that shares evenly into 6 piles, so count by sixes: 6, 12, 18, 24, 30, 36.
The first one that is 32 or more is 36.
So 36 - 32 = 4 more children are needed.

2024 · #12 Penguin Peter goes fishing every day and brings home 9 fish for his two children. Each day he gives 5 fish to the first child he sees,...

Penguin Peter goes fishing every day and brings home 9 fish for his two children. Each day he gives 5 fish to the first child he sees, and the other child gets the remaining 4 fish. Over the last few days, one child has received 26 fish in total. How many fish did the other child get?

Show answer

Answer: D — 28

Show hints

Hint 1 of 3

Every single day the two children together get 5 + 4 = 9 fish.

Still stuck? Show hint 2 →

Hint 2 of 3

Each day a child gets either 5 or 4, so try how many days it takes one child to reach 26.

Still stuck? Show hint 3 →

Hint 3 of 3

Once you know the number of days, the other child's total is 9 per day minus 26.

Show solution

Approach: find the number of days, then take the rest of the daily 9s

Each day one child gets 5 fish and the other gets 4, so together they get 9 fish a day.
For one child to reach 26 (made of 5s and 4s), it takes 6 days: 5 + 5 + 4 + 4 + 4 + 4 = 26.
In all, the children got 9 × 6 = 54 fish over those 6 days.
So the other child got 54 − 26 = 28 (D).

2013 · #11 Daniel had 36 sweets. He shared them equally between his siblings. How many siblings can he definitely not have?

Daniel had 36 sweets. He shared them equally between his siblings. How many siblings can he definitely not have?

Show answer

Answer: D — 5

Show hints

Hint 1 of 3

Sharing equally means the 36 sweets must split into equal groups with none left over.

Still stuck? Show hint 2 →

Hint 2 of 3

Try sharing 36 sweets among each number of siblings and see if any are left over.

Still stuck? Show hint 3 →

Hint 3 of 3

The number that leaves some sweets left over is the one he cannot have.

Show solution

Approach: test each option for dividing 36

36 divides evenly by 2, 3, 4 and 6 (giving 18, 12, 9, 6).
But 36 ÷ 5 is not a whole number.
So he definitely cannot have 5 siblings.

CHAPTER 4

Number Patterns

THEORY

Numbers march in patterns. Color every spot you land on, and a picture shows up like magic.

Here is a row of squares. Color every 3rd one:

The green spots are 3, 6, 9. They are the multiples of 3. The pattern keeps going the same way: 12, 15, 18...

Here is why patterns are your friend. Once you know one spot, you do not start over from 1. You just add the hop again. After 9 comes 12, then 15 — always 3 more than the last.

🎯 Try it

Color every 3rd spot: 3, 6, 9, ? — which spot is next?

Here is how: Each green spot is 3 more than the last. 9 plus 3 is 12.

For big kids: a digit-sum trick for 3

Add up the digits. If the sum is in the 3-times pattern (3, 6, 9, 12...), the whole number is too. Like 12: 1 + 2 = 3. Like 27: 2 + 7 = 9. Try it on 18 and see!

THE TRICK

THE MOVE: To find the next number in a skip-count, add the hop size again. Hopping by 3? Add 3 to the last one. Hopping by 5? Add 5.

WATCH OUT

Do not start counting from 1 every time. Once you know the last spot, just add the hop. Starting over is slow and easy to mess up.

WORKED EXAMPLE

PROBLEM · 2016 #9

If you add up the digits of the year 2016 (2 + 0 + 1 + 6), the result is 9. What is the next year after 2016 for which the sum of the digits is 9 again?

A) 2007 B) 2025 C) 2034 D) 2108 E) 2134

The digits of 2016 add up to 2 + 0 + 1 + 6 = 9. We want the next year whose digits also add to 9.

Check the years right after. 2017 adds to 10. 2018 adds to 11. The sums keep climbing past 9, not back to it.

Hop ahead and try 2025: 2 + 0 + 2 + 5 = 9. That works!

So the next year is 2025.

I add the four digits of each year and hunt for a 9. The years just after 2016 give sums that are too big, so I jump ahead until the digits add to 9 again.

Answer: B — 2025

RULE OF THUMB

A pattern repeats. Find the hop, then keep adding it to leap to the next spot.

MORE LIKE THIS

2021 · #11 Julie and Angela played “kangball”, a ball game. Each goal in their game scores 2 points. Julie scored 5 goals and Angela scored 9...

Julie and Angela played “kangball”, a ball game. Each goal in their game scores 2 points. Julie scored 5 goals and Angela scored 9 goals. How many more points than Julie did Angela score?

Show answer

Answer: C — 8

Show hints

Hint 1 of 3

First find how many MORE goals Angela scored than Julie.

Still stuck? Show hint 2 →

Hint 2 of 3

Each goal is worth 2 points, not 1.

Still stuck? Show hint 3 →

Hint 3 of 3

Turn just the extra goals into points.

Show solution

Approach: find the extra goals, then turn them into points

Angela scored 9 − 5 = 4 more goals than Julie.
Each goal is 2 points, so 4 goals is 4 × 2 = 8 points.
Angela scored 8 more points.

2015 · #4 If you multiply both digits of the number 35, you get 15. How big is the sum of both digits?

If you multiply both digits of the number 35, you get 15. How big is the sum of both digits?

Show answer

Answer: E — 8

Show hints

Hint 1 of 2

The two digits of 35 are 3 and 5; the problem already multiplies them.

Still stuck? Show hint 2 →

Hint 2 of 2

The question asks for the sum of those same two digits, not the product.

Show solution

Approach: read off the digits, then add them

The digits of 35 are 3 and 5 (and indeed 3 × 5 = 15).
Their sum is 3 + 5 = 8.
So the answer is 8.

CHAPTER 5

Tens and Ones (Place Value)

THEORY

Look at 27. It is not just a 2 and a 7 sitting side by side. The 2 means 2 tens. The 7 means 7 ones.

Where a digit sits changes how much it is worth:

Now swap the cards. 72 means 7 tens and 2 ones. Same two cards, but 72 is way bigger than 27! The card in front (on the left) is worth a whole lot more.

So to make a number BIG, put the big digit in front. To make it small, put the small digit in front.

🎯 Try it

You have two digit cards: 5 and 2. Put them together to make the BIGGEST two-digit number you can. What is it?

Here is how: Big digit in front. 5 is bigger than 2, so 5 goes first. That makes 52. (And 25 would be the smallest.)

THE TRICK

THE MOVE: The front card counts the most. Big digit in front makes the biggest number. Small digit in front makes the smallest.

WATCH OUT

A 0 cannot go in front of a number. 0 then 6 is not a real two-digit number — it is just 6. Keep zeros out of the front seat.

WORKED EXAMPLE

PROBLEM · 2014 #2

Marie wants to put the digit 3 somewhere into the number 2014. Where must she put the 3 so that the new number (with all 5 digits) is as small as possible?

A) in front of 2014 B) between 2 and 0 C) between 0 and 1 D) between 1 and 4 E) after 2014

Marie wants to slip a 3 into 2014 to make the smallest 5-digit number she can. There are 5 spots the 3 could go. Try each one and read it like a race:

In front: 32014. Between 2 and 0: 23014. Between 0 and 1: 20314. Between 1 and 4: 20134. After: 20143.

To pick the smallest, line them up and look at the front. The ones starting 20... are smaller than 32... or 23... So we keep 20314, 20134, 20143.

Among those, the third card decides: 20134 beats 20314, and 20134 is smaller than 20143. So the 3 goes between the 1 and the 4, making 20134.

Small numbers want small digits up front. The front of 2014 is already tiny (2, then 0), so I keep that and tuck the 3 near the back. Reading them like a race, 20134 wins.

Answer: D — between 1 and 4

RULE OF THUMB

The card in front is worth the most. Want it big? Big digit in front. Want it small? Small digit in front.

MORE LIKE THIS

2015 · #4 If you multiply both digits of the number 35, you get 15. How big is the sum of both digits?

If you multiply both digits of the number 35, you get 15. How big is the sum of both digits?

Show answer

Answer: E — 8

Show hints

Hint 1 of 2

The two digits of 35 are 3 and 5; the problem already multiplies them.

Still stuck? Show hint 2 →

Hint 2 of 2

The question asks for the sum of those same two digits, not the product.

Show solution

Approach: read off the digits, then add them

The digits of 35 are 3 and 5 (and indeed 3 × 5 = 15).
Their sum is 3 + 5 = 8.
So the answer is 8.

2012 · #19 Gregory made two 3-digit numbers from the digits 1, 2, 3, 4, 5, 6. Each digit was used only once. Afterwards he added the two numbers...

Gregory made two 3-digit numbers from the digits 1, 2, 3, 4, 5, 6. Each digit was used only once. Afterwards he added the two numbers together. What is the largest answer he could have got?

Show answer

Answer: D — 1173

Show hints

Hint 1 of 2

To make a sum large, put the biggest digits where they count the most.

Still stuck? Show hint 2 →

Hint 2 of 2

Give the two hundreds places the largest digits, then the tens, then the units.

Show solution

Approach: place the largest digits in the highest places

Use 6 and 5 in the hundreds places, 4 and 3 in the tens, 2 and 1 in the units.
The two numbers add to (600 + 500) + (40 + 30) + (2 + 1).
That is 1100 + 70 + 3 = 1173.
The largest possible sum is 1173.

★ MINI-QUIZ

Quick Check: Digit Cards

Two warm-ups about tens, ones, and making numbers. Remember: big digit in front!

2014 · #2 Marie wants to put the digit 3 somewhere into the number 2014. Where must she put the 3 so that the new number (with all 5 digits) is as...

Marie wants to put the digit 3 somewhere into the number 2014. Where must she put the 3 so that the new number (with all 5 digits) is as small as possible?

Show answer

Answer: D — between 1 and 4

Show hints

Hint 1 of 3

A number is smaller when its first (left-most) digits are smaller.

Still stuck? Show hint 2 →

Hint 2 of 3

Putting the big digit 3 near the front pushes the number up, so push the 3 as far right as you can.

Still stuck? Show hint 3 →

Hint 3 of 3

Write out each new number and read them like words to see which comes first.

Show solution

Approach: keep the small left-hand digits and push the 3 to the right

Try the 3 in each gap: 32014, 23014, 20314, 20134, 20143.
Reading them like a race, the one that stays smallest the longest at the front is 20134.
So the 3 should go between the 1 and the 4.
Answer: between 1 and 4.

2015 · #4 If you multiply both digits of the number 35, you get 15. How big is the sum of both digits?

If you multiply both digits of the number 35, you get 15. How big is the sum of both digits?

Show answer

Answer: E — 8

Show hints

Hint 1 of 2

The two digits of 35 are 3 and 5; the problem already multiplies them.

Still stuck? Show hint 2 →

Hint 2 of 2

The question asks for the sum of those same two digits, not the product.

Show solution

Approach: read off the digits, then add them

The digits of 35 are 3 and 5 (and indeed 3 × 5 = 15).
Their sum is 3 + 5 = 8.
So the answer is 8.

CHAPTER 6

Digits That Add Up

THEORY

Some puzzles hide the pieces and tell you only the total. Your job is to find the pieces that fit.

Like this: which two digit cards add up to 9?

When you know the total, walk through the pieces in order so you never miss one: 0 and 9, then 1 and 8, then 2 and 7, then 3 and 6, then 4 and 5. Tidy and complete.

And when the choices are given to you, the move is even simpler: try each choice in turn and keep only the ones that hit the total.

🎯 Try it

Two digit cards add up to 10. One of them is a 4. What is the other one?

Here is how: You need 10 in all. You already have 4. Count up from 4 to 10: that is 6 more. So the other card is 6.

THE TRICK

THE MOVE: When a puzzle gives you a total, try the choices in order. Add each one up. Keep the ones that match the total — toss the rest.

WATCH OUT

Some totals just cannot be made. If your coins only reach a certain amount, anything below that is impossible — not every number can be built.

WORKED EXAMPLE

PROBLEM · 2011 #11

Johannes has only 5 Cent coins and 10 Cent coins in his pocket. Altogether he has 13 coins. Which of the following amounts cannot be the total of his coins?

A) 80 c B) 60 c C) 70 c D) 115 c E) 125 c

Johannes has 13 coins, all 5 c or 10 c. We hunt for the total he can NOT make.

Find his smallest possible total first. If ALL 13 coins are 5 c, that is the least money he can have. Skip-count 13 fives: that reaches 65 c. He can never go below 65 c with 13 coins.

Now check the choices. 80, 70, 115, and 125 are all 65 or more — he can reach those by swapping some 5s for 10s.

But 60 c is below 65 c. With 13 coins that little is impossible. So 60 c cannot be his total.

I find the smallest he could have (all 5 c coins, which is 65 c). Any choice under that is impossible. 60 is under 65, so 60 is the odd one out.

Answer: B — 60 c

RULE OF THUMB

Given a total, walk the choices in order and test each. For a not-possible question, the ones that cannot be built are your answer.

MORE LIKE THIS

2010 · #24 Berti’s friends each add together the day and the month of their birthday. They all get the answer 35, but no two of them have the same...

Berti’s friends each add together the day and the month of their birthday. They all get the answer 35, but no two of them have the same birthday. What is the largest number of friends Berti can have?

Show answer

Answer: B — 8

Show hints

Hint 1 of 2

You need months and days with month + day = 35, and each birthday must be a real date.

Still stuck? Show hint 2 →

Hint 2 of 2

Start from December and step down through the months, checking the day fits that month.

Show solution

Approach: count valid (month, day) pairs summing to 35

List month + day = 35 with a valid day: Dec 23, Nov 24, Oct 25, Sep 26, Aug 27, Jul 28, Jun 29, May 30.
April would need day 31, which doesn't exist, and earlier months need impossible days.
That gives 8 different birthdays, so at most 8 friends.

2024 · #17 Mia has 3 cards, each showing a three-digit number. When she adds the three numbers she gets 782. Sadly a worm has eaten one digit on...

Mia has 3 cards, each showing a three-digit number. When she adds the three numbers she gets 782. Sadly a worm has eaten one digit on each card, so they now read 2 ? 3, 1 ? 4 and 4 1 ?. What do you get when you add the three digits the worm ate?

Figure for Math Kangaroo 2024 Problem 17

Show answer

Answer: D — 11

Show hints

Hint 1 of 3

First add up only the digits you can still see, putting a 0 in each eaten spot.

Still stuck? Show hint 2 →

Hint 2 of 3

Compare that total with 782 to see how much the eaten digits must add back.

Still stuck? Show hint 3 →

Hint 3 of 3

Remember: an eaten digit in a tens place is worth that many tens, and an eaten digit in a ones place is worth that many ones.

Show solution

Approach: add the visible digits first, then see how much the eaten digits must add back

Treat each eaten spot as 0: the cards read 203, 104 and 410, which add to 717.
But the real total is 782, so the eaten digits must add back 782 − 717 = 65.
Two of the eaten digits sit in tens places, so together they are worth 60 (meaning those two digits add to 6); the third sits in a ones place worth 5 (so that digit is 5).
The three eaten digits therefore add to 6 + 5 = 11 (D).

Another way:

⬢ FINAL TEST

Mini Test: Even, Odd & Skip-Counting

Seven real Kangaroo problems, easy first. Draw dots or hop along if it helps!

2013 · #11 Daniel had 36 sweets. He shared them equally between his siblings. How many siblings can he definitely not have?

Daniel had 36 sweets. He shared them equally between his siblings. How many siblings can he definitely not have?

Show answer

Answer: D — 5

Show hints

Hint 1 of 3

Sharing equally means the 36 sweets must split into equal groups with none left over.

Still stuck? Show hint 2 →

Hint 2 of 3

Try sharing 36 sweets among each number of siblings and see if any are left over.

Still stuck? Show hint 3 →

Hint 3 of 3

The number that leaves some sweets left over is the one he cannot have.

Show solution

Approach: test each option for dividing 36

36 divides evenly by 2, 3, 4 and 6 (giving 18, 12, 9, 6).
But 36 ÷ 5 is not a whole number.
So he definitely cannot have 5 siblings.

2017 · #13 13 children registered for a competition. Then another 19 joined. Six equally big teams are needed for the competition. How many more...

13 children registered for a competition. Then another 19 joined. Six equally big teams are needed for the competition. How many more children are needed, so that six equally big teams can be formed?

Show answer

Answer: D — 4

Show hints

Hint 1 of 2

First add the two groups of children together.

Still stuck? Show hint 2 →

Hint 2 of 2

Six equal teams means the total must split into 6 equal piles, so count up by sixes past your total.

Show solution

Approach: count up by sixes to the first number past the total

Altogether there are 13 + 19 = 32 children.
Six equal teams need a total that shares evenly into 6 piles, so count by sixes: 6, 12, 18, 24, 30, 36.
The first one that is 32 or more is 36.
So 36 - 32 = 4 more children are needed.

2011 · #9 A chicken farmer packs eggs in boxes of 6 and boxes of 12. What is the smallest number of boxes he needs to pack 66 eggs?

A chicken farmer packs eggs in boxes of 6 and boxes of 12. What is the smallest number of boxes he needs to pack 66 eggs?

Show answer

Answer: B — 6

Show hints

Hint 1 of 2

Use as many big boxes (of 12) as you can first.

Still stuck? Show hint 2 →

Hint 2 of 2

After the big boxes, fill the rest with boxes of 6.

Show solution

Approach: greedy with the larger box

66 = 12 × 5 + 6, so five boxes of 12 hold 60 eggs and one box of 6 holds the last 6.
That is 5 + 1 = 6 boxes, and no smaller number works, answer B.

2016 · #9 If you add up the digits of the year 2016 (2 + 0 + 1 + 6), the result is 9. What is the next year after 2016 for which the sum of the...

If you add up the digits of the year 2016 (2 + 0 + 1 + 6), the result is 9. What is the next year after 2016 for which the sum of the digits is 9 again?

Show answer

Answer: B — 2025

Show hints

Hint 1 of 2

The next year must be after 2016 and have digits adding to 9.

Still stuck? Show hint 2 →

Hint 2 of 2

Try years just after 2016 and add their digits.

Show solution

Approach: check years after 2016 for digit sum 9

Add the digits of each year right after 2016: 2017 gives 10, 2018 gives 11, and they keep climbing, so none of 2017–2024 lands back on 9.
Keep going to 2025: 2 + 0 + 2 + 5 = 9.
So the next such year is 2025, choice B.

2021 · #21 A box has fewer than 50 cookies in it. The cookies can be divided evenly between 2, 3, or 4 children. However, they cannot be divided...

A box has fewer than 50 cookies in it. The cookies can be divided evenly between 2, 3, or 4 children. However, they cannot be divided evenly between 7 children, because 6 more cookies would be needed. How many cookies are there in the box?

Show answer

Answer: D — 36

Show hints

Hint 1 of 2

Sharing evenly between 2, 3 and 4 children means the number is in the 2, 3 and 4 times tables — so it is in the 12 times table.

Still stuck? Show hint 2 →

Hint 2 of 2

List the multiples of 12 under 50, then check which one becomes a multiple of 7 after you add the 6 missing cookies.

Show solution

Approach: list multiples of 12 under 50, then test the sharing-by-7 clue

To share evenly between 2, 3 and 4 children the number must be in all three times tables, which means the 12 times table: 12, 24, 36, 48.
Needing 6 more cookies to share between 7 means that number plus 6 lands in the 7 times table.
Check each: 12+6=18, 24+6=30, 36+6=42, 48+6=54 — only 42 is in the 7 times table (6 x 7), so the number is 36.
So there are 36 cookies (D).

2012 · #19 Gregory made two 3-digit numbers from the digits 1, 2, 3, 4, 5, 6. Each digit was used only once. Afterwards he added the two numbers...

Gregory made two 3-digit numbers from the digits 1, 2, 3, 4, 5, 6. Each digit was used only once. Afterwards he added the two numbers together. What is the largest answer he could have got?

Show answer

Answer: D — 1173

Show hints

Hint 1 of 2

To make a sum large, put the biggest digits where they count the most.

Still stuck? Show hint 2 →

Hint 2 of 2

Give the two hundreds places the largest digits, then the tens, then the units.

Show solution

Approach: place the largest digits in the highest places

Use 6 and 5 in the hundreds places, 4 and 3 in the tens, 2 and 1 in the units.
The two numbers add to (600 + 500) + (40 + 30) + (2 + 1).
That is 1100 + 70 + 3 = 1173.
The largest possible sum is 1173.

2010 · #24 Berti’s friends each add together the day and the month of their birthday. They all get the answer 35, but no two of them have the same...

Show answer

Answer: B — 8

Show hints

Hint 1 of 2

You need months and days with month + day = 35, and each birthday must be a real date.

Still stuck? Show hint 2 →

Hint 2 of 2

Start from December and step down through the months, checking the day fits that month.

Show solution

Approach: count valid (month, day) pairs summing to 35

List month + day = 35 with a valid day: Dec 23, Nov 24, Oct 25, Sep 26, Aug 27, Jul 28, Jun 29, May 30.
April would need day 31, which doesn't exist, and earlier months need impossible days.
That gives 8 different birthdays, so at most 8 friends.

APPENDIX

Quick reference

Memorize these

Even ends in 0, 2, 4, 6, 8. Odd ends in 1, 3, 5, 7, 9.
Multiples of 5 end in 5 or 0. Multiples of 10 end in 0.
The leftover is always smaller than the number of friends.
The digit in front is worth the most: big in front makes a big number.

→ Practice Number Theory problems